# Why does the voltage drop when powering, then reverse when disconnecting this electromagnet?

I've just bought a 12V 260mA electromagnet and I have some questions about it

1. I don't know why, but when I connect electromagnet, my 12V power source (9V and 2 1.5V AA batteries in series) drops to 8V. I suspect that this is because of the protection of the 9V battery, because the 9V battery drops to 5V whereas voltage drop of 2 1.5V AA batteries is negligible, and when I try different 9V battery, it keeps and slowly drops to 2V.

2. When I disconnect the electromagnet, it seems that it sends reverse voltage, Before I disconnect, my multimeter connected in reverse reads around -8V, but when I disconnect it, it reads around 2V. Since I'm gonna switch electromagnet using transistor, would this reverse voltage be harmful? If so, just connecting the diode would prevent this?

3. My original plan is using 2 9V batteries in series to create 18V and using a voltage divider to divide it into 6V and 12V, and using 6V to power the rest of the circuit. If the voltage drop stated in 1 is actually caused by the protection of 9V batteries, is there any way to bypass it?

• 9V batteries don't have "protection." 9V batteries are simply incapable of delivering high current. It's like putting a lawn mower engine in place of of the big diesel engine in a large truck, and then saying that the truck doesn't move because the lawn mower engine has some kind of power limiting device in it.
– JRE
Oct 20, 2020 at 16:14
• Don't keep the multimeter connected when disconnecting the coil. The inductive kickback has extremely high voltage and could damage the multimeter. Also, forget 9V batteries powering this. Oct 20, 2020 at 18:13

Your plan of using 9v "transistor radio" batteries is not workable for this purpose.

Probably the minimum suitable would be a series collection of some number of AA cells. It's better that you use conventional flashlight batteries and not lithium for such a learning process.

A voltage divider is not really suitable either, rather tune the number of cells to fit your need. If you really wanted power adjustment you'd probably use PWM controlled by an MCU.

Your coil is an inductive load so if electronically switching it, you need to put a diode "backwards" across the coil to provide a path for the inductive kick at turnoff.

You may wish to search and read the various questions here on controlling a solenoid or relay coil with a transistor, as that is basically what you are also doing.

• Here's a fun trick. Take the bare end of each magnet wire between your fingers, then touch the wires to the battery for a few seconds and hold it there. Then pull the wire away from the battery. Well, it's fun to do to other people. It demonstrates why you need a diode.. Oct 20, 2020 at 22:20
• I should've noted that I'm using a small electromagnet, that only requires 12v 260mA and can lift up 2.5KG, which isn't that powerful. The reason i used 9v battery is the electromagnet is so small that it doesn't seem to require that high power, and i only need it to be able to lift around 500g. Some strange behavior i just discovered is, when I connect my 9v battery to small dc motor that can run steadily on 3v, its output voltage decreases to under 5v Oct 22, 2020 at 8:53
• I measured again and 9v battery drops to 2.3v when i connect to the small dc motor Oct 22, 2020 at 9:04
• Yes, that's why this is saying to use AA's bare minimum, better C"s or D's. Oct 22, 2020 at 9:46

I don't know why, but when I connect electromagnet, my 12v power source(9v and 2 1.5V AA batteries in series)drops to 8v

9V batteries are not meant to supply large currents. Their voltages drop under high current draw. You picked the wrong battery type.

When I disconnect the electromagnet, it seems that it sends reverse voltage,

Flyback voltage. You need to remind yourself of the basic lessons about inductors (which your electromagnet is an example of)!

Since I'm gonna switch electromagnet using transistor, would this reverse voltage be harmful?

Yes.

if so, just connecting the diode would prevent this?

Which diode? If you mean a flyback diode: yes, an appropriately dimensioned flyback diode might help.

I feel like you've not really understood the time behaviour of inductors, but that's really part of the basics of linear circuits – you might want to revisit the basics of electronics before dealing with semiconductors like transistors.

My original plan is using 2 9V batteries in series to create 18v and using a voltage divider to divide it into 6v and 12v, and using 6v to power the rest of the circuit.

Terrible plan. A voltage divider is not a power supply. And still, 9V batteries: wrong choice.

• Loading up your batteries will cause the voltage at the terminals to drop due to the internal resistance of the batteries. Like Marcus said, the 9V battery isn't meant to supply relatively large currents like an electromagnet so that's why you're seeing such a large drop. The reverse voltage is what inductors do when you give them a sudden change in input current and yes, you will need a diode to pass the current safely and prevent damaging the rest of your circuit. As for powering the rest of the circuit, DC-DC converters are cheap and take out the worries about changing battery voltage.
– vir
Oct 20, 2020 at 15:54
1. The 9V battery can't supply the current you need for the electromagnet, at least not for a long time; it is too small. The voltage will drop almost immediately, and the battery won't last long.

2. The electromagnet has a coil, an inductance, and there will be a kickback when you disconnect it. You will need a diode you mention to protect the rest of the circuitry.

3. Not a good plan; voltage dividers don't make good power supplies, and neither do 9 V batteries for anything else than very small currents.